So if you forget the coefficients you can rebuild them, and the symmetry stops being a separate fact to learn.
Pascal's triangle, built one row at a time. Each entry is the two above it added. Build it and watch row 5 appear.
Row n gives the coefficients of (a + b)ⁿ.
(a + b)n = ∑ nCr an−r br, with r running from 0 to n. The powers of a count down while the powers of b count up, and they always add to n, which is the fastest check on any term you write.
Substituting x = 1 checks every coefficient at once. The expansion must equal (1 + 2)⁵ = 243, and if your six numbers do not total 243, one of them is wrong. It costs five seconds.
Most questions ask for a single term, so expanding everything wastes time.
The r is not the power you want. Here you want x² and r is 3, because r counts the other bracket. Writing "powers add to n" first, every time, is what stops this.
1. Find 5C2.
2. Find the coefficient of x³ in (x + 2)⁵.
3. What do the entries of row 5 add up to?
Writing the general term with its nCr before substituting anything. It is a method mark even if the arithmetic then slips.
Both powers present and adding to n, including the one that is often left off when the bracket is (x + 2) rather than (a + b).
Raising the whole of the second term, so (2x)³ is 8x³ and not 2x³.
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